course Phy 202
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23:17:48 Query problem set 1 #'s 17-24 If we know the initial KE of a particle, its charge and the uniform electric field in which it moves, then if the net force on the particle is due only to the electric field, how do we find the KE after the particle has moved through a given displacement?
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RESPONSE --> First we would use `dW + `dKE = 0, the Work-Energy Theorem and then we could use .5 m v^2 = KE0 - `dW
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23:19:27 ** GOOD STUDENT SOLUTION: Given KE0, q, E, `ds: First we can find the Force by the relationship, q*E. Next, we can use the Force found to find the work done: `dW = F * `ds By the relationship `dW +`dKE = 0, we can then find `dKE, which we combine with KE0 to get KEf. **
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RESPONSE --> Ok this is what i found
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23:22:35 If we know the charge transferred between two points, the time and the average power necessary to accomplish the transfer, how do we find the potential difference between the points?
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RESPONSE --> First find the force on the charge is F = q * `dV / `dx and then the work done `dW = F * `dx = q * `dV / `dx * `dx which leads you to the average rate at which work is done is P =q `dV / `dt.
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23:23:48 ** The potential difference is found from the work done and the charge. Potential difference, or voltage, is work / charge, in Joules / Coulomb. We find the work from the power and the time, since power = work / time. **
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RESPONSE --> ok
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23:26:33 Explain how we can use the flux picture to determine the electric field due to a point charge Q at a distance r from the charge.
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RESPONSE --> The flux of a charge Q is 4 `pi k Q. When spread out over a sphere of radius r, the flux density is 4 `pi k Q / (area of the sphere) = 4 `pi k Q / (4`pi r^2) = k Q / r^2. This is the electric field strength. At distance 2r, the field is similarly found to be k Q / (2r)^2 = (1/4) k Q / r^2, or 1/4 as great as at distance r. If charge is doubled the field will be k (2Q) / r^2, twice as great as for the original charge at this distance. If distance and charge are both doubled the field becomes k (2Q) / (2r)^2 = (1/2) k Q / r^2, half as great as for the original charge and distance
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23:26:52 STUDENT RESPONSE AND INSTRUCTOR COMMENT: Flux = 4pikQ Flux = area of sphere * electric field = 4 pi r^2 * E k is 9.0 x 10^9 N m^2/C^2 We have 4 pi r^2 * E = 4 pi k Q so E = 4 pi k Q / ( 4 pi r^2) = k Q / r^2 INSTRUCTOR COMMENT: ** Note that the sphere is centered at the charge Q and passes thru the point at distance r so the radius of the sphere is r. Note also that this works because the electric field is radial from Q and hence always perpendicular to the sphere. **
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RESPONSE --> I see that the sphere is centered at the charge Q and passes thru the point at distance r so the radius of the sphere is r.
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23:29:03 Explain how we can use the flux picture to determine the electric field due to a charge Q uniformly distributed over a straight line of length L, at a distance r << L from that line but not close to either end.
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RESPONSE --> Q will produce flux 4 `pi k Q electric field = flux / area = 2 k Q / (r L).
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23:29:15 ** imagine a circular cylinder around a long segment of the wire; determine the charge on the segment. Total flux is 4 pi k * charge. By the symmetry of the situation the electric field has a very nearly constant magnitude over the curved surface of the cylinder (for an infinite wire the field would be absolutely constant). Almost all of the flux exits the curved surface of the cylinder and is at every point perpendicular to this surface (for an infinite wire all the flux would exit thru the curved surface and would be exactly perpendicular). So you can find flux / area, which is the field. You get E = flux / area = 4 pi k Q / ( 2 pi r * L) = 2 k Q / L. **
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RESPONSE --> Exactly
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23:29:36 Why does the charge on the initially uncharged capacitor build more and more slowly if we continue cranking the generator at a constant rate?
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RESPONSE --> It stores some of the energy
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23:30:05 ** The generator supplies a constant voltage; the charge flowing through the circuit builds on the capacitor, which builds voltage opposed to that of the generator. This results in less voltage across the rest of the circuit and therefore less flow of current. Less current requires less energy so the cranking is easier. UNIV PHY NOTE: Think of charge building on the capacitor, not current. The capacitor actually stores up the integral of the current with respect to time; since current is charge / time this integral has units of charge. **
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RESPONSE --> OK, so there is less voltage across the rest of the circuit and therefore less flow of current
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